RESULTADOS

P03 01

Equations

3.1 Calculation of blackbody emissive power Arrays are used but a Parametric Table could have been used.

We can use the EES built-in constants, C1# and C2#

λ₁= 10; T₁= 1000 (1)

To show all variables in the Solutions Window use:

$Arrays off

The results are consistent with Figure 3.4.1.

Solution

P03 02

Equations

3.2 Determination of the fraction of blackbody radiation in a wavelength range

This can be done by computing the fraction of the energy in the spectrum from zero to the two wavelengths with Equation 3.6.4 and taking the difference. Table 3.6.1a could have been used to find the two fractions.

T = 300 [K] (1)

λ₁= 8 [micrometer] (2)

λ₂= 14 [micrometer] (3)

f2= BlBodyF ract (λ2· T ) EES function or Eqn 3.6.4 (4)

f1= BlBodyF ract (λ1· T ) EES function or Eqn 3.6.4 (5)

f = f2− f1 (6)

Solution

f =0.3757 f₁= 0.1402 f₂= 0.516 λ₁= 8[micrometer]

λ₂= 14[micrometer] T = 300[K]

P03 03

Equations

3.3 We could use EES to integrate Eqn 3.6.3 or we could use Eqn 3.6.4 and sum for 10 terms.

The second method is used for the SETP EES functions. Here we will do the integration using the EES integral function.

SubprogramBBf (LT 1, LT 2 : f ) (1)

f = (c1#/sigma#) · Z LT 2

LT 1

 1

LT⁵· (exp (C2#/LT ) − 1)



dLT (2)

end (3)

LT 1 = 1000 (4)

LT 2 = 5000 (5)

call BBf (LT 1, LT 2 : f ) (6)

For the given conditions the table gives f=0.6334 and the integral gives 0,6335.

Note that a subprogram was not necessary but was requested in the problem statement. The solution could have been:

LT1=1000 LT2=5000

f=(c1#/sigma#)*Integral(1/(LT⁵*(exp(C2#/LT)-1)), LT, LT1, LT2)

Solution

Variables in Subprogram BBf

f = 0.6335 LT = 5000[micrometer] LT 1 = 1000[micrometer] LT 2 = 5000[micrometer]

76

P03 04

Equations

3.4 Calculation of radiant energy transport between large parallel plates:

T₁= 500; ₁= 0.45; T₂= 300; ₂= 0.20; σ = sigma#

qrad= σ · T₁⁴− T₂⁴

(1/₁+ 1/₂− 1) Eqn 3.8.4 qrad= hrad· (T1− T2) Eqn 3.10.1

Solution

1= 0.45 2= 0.2 hrad=2.478W/m²-K

qrad=495.7W/m² σ = 5.670 × 10⁻⁸W/m²-K⁴

T1= 500[K] T2= 300[K]

P03 05

Equations

3.5 Calculation of a heat loss coefficient for combined wind and radiation loss for a plate facing up.

Tamb= ConvertTemp(C, K, 10) T_{P late}= ConvertTemp(C, K, 50) T_{DewP oint}= 3 [C]

T ime = 0

 = 0.88

hwind= 25 W/m²·K

Estimate an equivalent sky temperature using the Berdahl & Martin method, Equation 3.9.2. Note the unusual units in this Equation.

T_sky = T_amb·

0.711 + 0.0056 [1/C] · T_{DewP oint}+ 0.000073 1/C² · (TDP)²+ 0.013 · Cos(15 [deg/hr] · T ime)^.25

Radiation heat transfer is given by Equation 3.9.1. Divide this heat transfer by (TP late- Tamb) to get the radiation coefficient based on the ambient temperature.

h_Rad= sigma# ·  · T_{P late}⁴ − T_sky⁴ TP late− Tamb

Add the convection coefficient to get the total loss coefficient.

U_overall= h_Rad+ h_wind

Solution

 = 0.88 hRad= 7.657W/m²-K

hwind= 25W/m²-K

T ime = 0[hr]

Tamb= 283.1[K] TDewP oint= 3[C] TP late= 323.1[K] Tsky = 262.7[K]

Uoverall=32.7W/m²-K

78

P03 06

Equations

3.6 Calculation of convection heat transfer in a collector.

The Rayleigh number is found from Eqn 3.11.2 Ra = g# · beta⁰· (T1− T2) · l³

µ · α

The Nusselt Number is found from Eqn 3.11.4

N us = 1 + 1.44 · 1 − 1708 · (sin (1.8 · min (Slope, 75)))^1.6 Ra · Cos(min (Slope, 75))

!

· A + B

where A and B must be constrained to be positive or zero and for slopes greater than 75, use slope = 75:

A = max

 0,



1 − 1708

Ra · Cos(min (Slope, 75))



B = max 0,



Ra ·Cos(min (Slope, 75)) 5830

The three conditions are done in a Parametric Table.

Table 1

Run l Slope N us Ra hconv Qconv

[m] [deg] [-] [-] W/m²-K

W/m²

P03 07

Equations

3.7 Calculation of equilibrium temperatures of plates exposed to solar radiation

a For a plate in space with surface normal to the solar radiation an energy balance on the plate (which absorbes solar radiation on one side and emits radiation from both) is:

Gsc#· αsolar = 2 · Sigma# ·  · T_plate,a⁴ From Table 4.7.1 the solar absorptance is:

α_solar = 0.35

Assume the emittance of copper, from Table 4.7.1,can be approximated by a parabola

 = a + b · Tplate,a+ c · T_plate,a²

0.041 = a + b · 338 [K] + c · 338 [K]²; 0.036 = a + b · 463 [K] + c · 463 [K]²; 0.039 = a + b · 803 [K] + c · 803 [K]²

b For a plate in the atmosphere with surface normal to the solar radiation and a zenith angle of 90 deg an energy balance on the plate (which absorbes solar radiation on one side, exchanges radiation from the top side with the sky and from the bottom side with the ground, and loses energy by convection from both sides to the atmosphere) is:

b = a + b · Tplate,b+ c · T_plate,b²

Gsc#· τatm· αsolar= Sigma# · b· T_plate,b⁴ − T_sky⁴
+ T_plate,b⁴ − T_ground⁴

+ 2 · hwind· (Tplate,b− Tamb) τatm= 0.8; Tamb = 298 [K] ; Tsky = 288 [K] ; hwind= 23 W/m²·K ; Tground= 288 [K]

The convection term in the energy balance dominates so it is clear that the exact value of the emittance is not critical.

Solution

a = 0.07095[-] αsolar= 0.35 b = −0.0001241[1/K] c = 1.050 × 10⁻⁷1/K²

 = 0.03428 b= 0.0428 hwind= 23W/m²-K

τatm= 0.8 Tamb= 298[K] Tground= 288[K] Tplate,a=592[K] Tplate,b=306[K]

T_sky = 288[K]

80

P03 08

Equations

3.8 Energy balance calculations on a hypothetical instrument for measuring solar radiation.

Knowns

hconv= 16 W/m²·K ; Tdisk,b = ConvertTemp(C, K, 15); Tdisk,w= ConvertTemp(C, K, 5); cover= 0.88

disk = 0.95; τcover= 0.90; αb = 0.95; αw= 0.35

a An energy balance on each disk, assuming that the disks and cover act like parallel plates for radiation exchange:

G · τ_cover· αb= 2 · Sigma# · T_disk,b⁴ − T_cover⁴

(1/disk+ 1/cover− 1) + h_conv· (Tdisk,b− Tcover)

!

G · τcover· αw= 2 · Sigma# · T_disk,w⁴ − T_cover⁴

(1/disk+ 1/cover− 1)+ hconv· (Tdisk,w− Tcover)

!

b Heat transfer coefficient from disk to cover:

c An expression for incident radiation

Incident solar energy, G, is expressed in terms of the temperature difference between black and white disks. Note that the first two equations are sufficient to solve for G; so the new variable G1 is introduced.

For the black disk:

d Calculation of incident radiation

In the days before EES, the nonlinear radiation term would have been linearized using the second two equations. The solution would then involve guessing the cover temperature, finding the U’s, substituting into the equation of part c and then checking the guess of the cover temperature using one of the commented equations of pary c.. With EES we do not need to worry about the solution technique - let EES do it. Also note that the ambient temperature was not used except maybe to get a reasonable guess of Tcover.

Solution

P03 09

Equations

3.9 Convection loss coefficient and convection heat transfer in a collector

$UnitSystem SI Mass Deg kPa K kJ

Tu= ConvertTemp(C, K, 110 [C]); Tl= ConvertTemp(C, K, 60 [C]); T =¯ T_l+ T_u

We can use the SETP functions to fing the Nusselt Number N u# = N uF latP late (Slope, T_u, T_l, L)

or we can evaluate the Rayleigh number and then find the Nusselt number.

Ra# = g# · β · ∆T · L³

SlopeL = min (Slope, 75 [deg]) Slope is constrained to be between 0 and 75 A = max

As expected, both Nusselt numbers are identical.

N u# = h · L/k; q = h · ∆T

P03 10

Equations

3.10 Effect of a slat-type honeycomb on the convection heat transfer of Problem 3.9

$UnitSystem SI Mass Deg kPa K kJ

T_u= ConvertTemp(C, K, 110 [C]); T_l= ConvertTemp(C, K, 60 [C]); T =¯ T_l+ T_u

From Figure 3.12.2 at an AspectRatio of 0.5:

C1= 0.113; C2= 1.00

We need to evaluate the Rayleigh number and then find the Nusselt number.

Ra# = g# · β · ∆T · L³

P03 11

Equations

Heat transfer coefficients for air flowing in a rectangular duct .

$UnitSystem SI Mass Deg kPa C kJ Knowns, properties and definitions

Length = 2.0 [m] ; width = 1 [m] ; Tair= 25 [C] ; µ = µ (air, T = Tair) ; P r# = Pr(air, T = Tair)

k = k (air, T = Tair) ; Af low = width · depth; P = 2 · (width + depth) ; Dh=4 · Af low

P m=ρ*A˙ f low*V and Re#=ρ*V*Dh/µ so that when ρ is eliminated we have

Re# = m · D˙ h

µ · Af low

Use a parametric table for the three situations. From Eqn 3.14.7 N u# = 4.9 + .0606 · (Re# · P r# · Dh/Length)^1.2

1 + .0909 · (Re# · P r# · D_h/Length)^.7· P r#^.17 N u# = h · Dh/k

For situation 3 the Reynolds number is in the transition region. For conservative design use the laminar flow results. The turblent heat transfer coefficient from Eqn 3.14.6 is 7.26 W/m²-K

Solution

A_{f low}= 0.015m²

depth = 0.015[m] D_h= 0.02956[m] h = 5.718W/m²-K

k = 0.02551[W/m-K]

Length = 2[m] µ = 0.00001849[kg/m-s] m = 0.024˙ [kg/s] N u# = 6.625[-] P = 2.03[m]

P r# = 0.7281 Re# = 2, 558 Tair= 25[C] width = 1[m]

Table 1

Run depth m˙ Re# ν# h

[m] [kg/s] [-] W/m²-K

1 0.015 0.012 1,279 5.815 5.019 2 0.0075 0.012 1,289 5.367 9.197 3 0.015 0.024 2,558 6.625 5.718

P03 12

Equations

3.12 Calculation of pressure drop in a packed bed using the Dunkle & Ellul correlation

$UnitSystem SI Mass Deg kPa K kJ C

Length = 2.0; M assF lowRate = 1.1 [kg/s] ; Area = 12 m² ; P artDiam = 0.02 [m]

T = 40 [C] ; DenAir = ρ (Air, T = T, p = 101.3) ; V isc = µ (air, T = T )

M assV el = M assF lowRate/Area

Equation 3.16.5 or use the SETP function for pressure drop.

∆p = P resDrop (Length, M assV el, P artDiam, T )

Solution

Area = 12m²

∆p = 29.31[Pa] DenAir = 1.127kg/m³

Length = 2[m]

M assF lowRate = 1.1[kg/s] M assV el = 0.09167kg/s-m²

P artDiam = 0.02[m] T = 40[C]

V isc = 1.918 × 10⁻⁵[kg/m-s]

86

P03 13

Equations

3.13 Estimation of particle size to get a pressure drop of 55 Pa for the circumstances of Problem 3.12

$UnitSystem SI Mass Deg kPa K kJ C

Note that with EES it is only necessary to delete the particle diameter in problem 3.12 and add the pressure drop.

Length = 2.0; M assF lowRate = 1.1 [kg/s] ; Area = 12 m² ; ∆p = 55 [Pa]

T = 40 [C] ; DenAir = ρ (Air, T = T, p = 101.3) ; V isc = µ (air, T = T )

M assV el = M assF lowRate/Area

Equation 3.16.5 or use the SETP function for pressure drop.

∆p = P resDrop (Length, M assV el, P artDiam, T )

An alternative solution is to use Equation 3.16.4 by McCorquodale et al.

V F = 0.45

P03 14

Equations

3.14 Calculation of upward heat loss terms for a collector.

$UnitSystem SI Mass Deg kPa K kJ Data:

Tplate= ConvertTemp(C, K, 100); Tcover= ConvertTemp(C, K, 31.5); Tsurr= ConvertTemp(C, K, 10)

plate= 0.10; cover= 0.88; Slope = 50; σ = sigma#; g = g#; L = .025 [m]

a Radiation Heat Transfer Qrad= σ · T_plate⁴ − T_cover⁴

1/plate+ 1/cover− 1 b Radiation coefficient hrad= Qrad

T_plate− T_cover c Convection Heat Transfer

N u# = N uF latP late (Slope, T_cover, T_plate, L) Equation 3.11.14 T_mean= Tplate+ Tcover

2 k = k (Air, T = Tmean) N u# = h_conv· L/k

Qconv= hconv· (Tplate− Tcover)

Instead of using the NuFlatPlate EES function we could have used either Figure 3.11.1 or Eqn 3.11.4 along with the following:

β = 1/Tmean

Cp = cp(Air, T = Tmean)

88

ρ = ρ (Air, T = Tmean, P = 101.3)

d Radiation from cover to surroundings QradCover,Surr= cover· σ · T_cover⁴ − T_surr⁴

We can determine QconvCover,Surr since loss is either convection or radiation.

QconvCover,Surr = Qconv+ Qrad− QradCover,Surr

Solution

QconvCover,Surr = 198.9W/m²

Qrad= 60.26W/m²

p03 15

Equations

3.15 Estimation of wind convection coefficients

$UnitSystem SI Mass Deg kPa K kJ Data:

Length = 10 [M] ; W idth = 2.5 [M] ; V el = 5.0 [m/s]

T_Surr = ConvertTemp(C, K, 10); T_cover= ConvertTemp(C, K, 31.5)

Evaluate properties at the mean air tempeature.

Tprop= Tsurr+ Tcover

2 ; k = k (‘air’ , T = Tprop) ; ρ = ρ (‘air’ , T = Tprop, P = 101.3) µ = µ (‘air’ , T = Tprop) ; P r = Pr(air, T = Tprop)

a For a free standing colector:

LChar= 4 · Length · W idth

2 · (Length + W idth); Re = LChar· V el · ρ/µ N u#_wind= 0.86 · Re^1/2· P r^1/3 Eqn 3.15.1

N u#_wind= h_wind,a· L_char/k

Note: The Re is 1.31*10⁶, which is slightly beyond the range of the experiments on which Equation 3.15.1 is based. With no satisfactory alternative, use it.

b For a flush-mounted collector on a building

V olume = 564; h_wind,b= max 5 W/m²·K , 8.6 · V el^.6 V olume^1/3
.4

!

Eqn 3.15.10

Note: Equations 3.15.9 and 3.15.10 are not dimensionally correct unless the 8.6 has units of (W/m²-K)/(m^0.2/s^0.6). For empirical equations like this it is best to just turn off unit checking for that equation.

Solution

P03 16

Equations

3.16 Effectiveness-NTU calculations for a counterflow heat exchanger

$UnitSystem SI Mass Deg kPa K kJ Data:

˙

mH2O= 3.75 [kg/s] ; U A = 2.1 × 10⁵ [W/K] ; Tin,H2O= 49 [C] ; Tin,Glycol= 65 [C] ; Cp,Glycol = 3780 [J/kg ·K] ; Cp,H2O = 4190 [J/kg ·K]

Ccold= ˙mH2O· Cp,H2O

Chot= ˙mGlycol· Cp,Glycol

C_min= min (C_cold, C_hot) C_max= max (C_cold, C_hot) N T U = U A/C_min

C^∗= Cmin/Cmax

Ef f = 1 − exp (−N T U · (1 − C^∗)) 1 − C^∗· exp (−N T U · (1 − C^∗)) Q = Ef f · C_min· (T_in,Glycol− T_in,H2O) Q = Ccold· (Tout,H2O− Tin,H2O) Q = Chot· (Tin,Glycol− Tout,Glycol)

Run a parametric table with ˙mglycolreanging from 0 to 20. The solution is shown for the specific case of :

˙

mglycol= 5

Solution

C_cold= 15, 713[W/K] C_hot= 18, 900[W/K] C_max= 18, 900[W/K] C_min= 15, 713[W/K]

C_p,Glycol= 3, 780[J/kg-K] C_p,H2O= 4, 190[J/kg-K] C^∗ = 0.8313[-] Ef f = 0.981[-]

˙

mGlycol= 5.0[kg/s] m˙H2O= 3.75[kg/s] N T U = 13.37[-] Q = 246, 524[W]

Tin,Glycol= 65[C] Tin,H2O= 49[C] Tout,Glycol= 51.956419181[C] Tout,H2O= 64.69[C]

U A = 210, 000[W/K]

T

out

vs m

dotglycol

92

P03 17

Equations

3.17 Heat transfer from the hot absorber plate to the ambient.

$UnitSystem SI Mass Deg kPa K kJ Data:

_p = 0.2; _g= 0.88; slope = 60 [deg] ; h_wind= 7 W/m²·K ; T_amb= ConvertTemp(C, K, 15) (1)

P lateSp = 0.030 [m] ; σ = sigma#; T_sky= T_amb; T_plate= ConvertTemp(C, K, 85) (2)

a Heat loss from collector

We can use the SETP function for Eqn 3.11.4 to find hplate,cover

N u# = N uF latP late (Slope, T_cover, T_plate, P lateSp) (3)

N u# = hplate,cover· P lateSp/k (4)

k = k



air, T = Tcover+ Tplate

2



(5)

The various energy flows through the cover system are:

q_conv,p,c= hplate,cover· (T_plate− T_cover) (6)

q_rad,p,c= σ · T_plate⁴ − T_cover⁴

1/_p+ 1/_g− 1 (7)

q_wind= h_wind· (Tcover− Tamb) (8)

q_rad,sky = σ · _g· T_cover⁴ − T_sky⁴

(9)

The cover temperature is found from an energy balance on the cover.

qconv,p,c+ qrad,p,c= qrad,sky+ qwind (10)

The heat loss through the cover is then:

qloss= qconv,p,c+ qrad,p,c (11)

b Heat loss coefficient as a function of plate temperature is found by commenting out the equation above that sets the plate temperature to 85 C and setting the plate temperature (in C) in a parametric table.

Tplate= ConvertTemp(C, K, TplateC) (12)

qloss= U · (Tplate− Tamb) (13)

c What energy transfers are we neglecting in this analysis of a collector?

back losses through the insulation.

edge losses

two-dimensional losses in the corners.

Solution

U vs T

plate

P04 01

Equations

4.1 Calculation of equilibrium temprerature of a surface in space.

Assume that the 6000 K radiation is normal to the surface and that there is no loss other than radiation from the surface.

Tsun= 6000 [K] ; S = 1367 W/m²

Use a Parametric table to specify the critical wavelength.

ρ0,λ = 0.1; ρλ,∞= 0.9

Fraction of the sun’s energy between zero and λ

f_o,λ = BlBodyF ract (λ · T_sun) From the EES SETP library.

ρsun= fo,λ · ρ0,λ + (1 − fo,λ) · ρλ,∞

αsun= 1 − ρsun

Fraction of the IR radiation from surface at Tsurf ace

fo,λsurf = BlBodyF ract (λ · Tsurf)

ρ_surf = f_o,λ_surf· ρ0,λ + (1 − f_o,λ_surf) · ρ_λ,∞

α_surf = 1 − ρ_surf α_surf = _surf

Energy balance on surface assuming back is insulated.

αsun· S = surf· sigma# · T_surf⁴

If we did this by hand we would need to guess a surface temperature, find the fraction of the IR radiation in each band, find a temperature from the energy balance and repeat the process until the guess and calculated temperatures are equal.

Table 1

Run λ fo,lambda fo,lambda,surf αsun surf Tsurf

[micrometer] [K]

1 1 0.7378 3.322E-7 0.690 0.100 638.7

2 2 0.9451 0.005337 0.856 0.104 667.0

3 3 0.9808 0.049 0.88596 0.139 625.7

P04 02

Equations

4.2 Calculation of absorptance and emittance for a surface with a step change in spectral reflectance

$UnitSystem SI Mass Deg kPa K kJ Data:

Tsun= 5777 [K] ; Tsurf ace= ConvertTemp(C, K, 150) (1)

αλ,1= 0.95; αλ,2= 0.05; λ = 1.8 [micrometer] (2)

fraction of solar radiation below λ from Eqn 3.6.3.

f_sol= BlBodyF ract (λ · T_sun) (3)

From Eqn 4.6.4 for two bands

α = αλ,1· fsol+ αλ,2· (1 − fsol) (4)

fraction of IR radiation below λ from Eqn 3.6.3.

f_IR= BlBodyF ract (λ · T_{surf ace}) (5)

From Eqn 4.6.1 for two bands and using αλ= λfrom Eqn 4.4.5

 = αλ,1· fIR+ αλ,2· (1 − fIR) (6)

Solution

α = 0.880 αλ,1= 0.95 αλ,2= 0.05  = 0.050 fIR= 0.000007637 fsol= 0.9219 λ = 1.8[micrometer] Tsun= 5, 777[K]

Tsurf ace= 423.1[K]

p04 03

Equations

4.3 Calculation of absorptance and emittance of a surface from spectral reflectance data.

$UnitSystem SI Mass Deg kPa K kJ

a For the extraterrestrial solar spectrum. Using 10 equally spaced intervals from Table 1.3.1b, the wavelength at center points are: 0.364, 0.455, 0.525, 0.599, 0.682, 0.787, 0.923, 1.113, 1.412, 2.117. Reading corresponding reflectance values from figure 4.8.3

ρ₁= 0.04; ρ₂= 0.04; ρ₃= 0.04; ρ₄= 0.04; ρ₅= 0.04 ρ6= 0.04; ρ7= 0.05; ρ8= 0.06; ρ9= 0.10; ρ10= 0.32 αextraterrestrial= 1 − Sum(ρi, i = 1, 10)

10

b For the standard terrestrial solar spectrum. Using 10 equally spaced intervals from Table 2.6.1, the appropriate wavelenghts are: 0.426, 0.508, 0.581, 0.653, 0.732, 0.822, 0.929, 1.080, 1.300, 1.974. Note that the numbers differ little from part a. Reading reflectance values from figure 4.8.3:

r1₁= 0.04; r1₂= 0.04; r1₃= 0.04; r1₄= 0.04; r1₅= 0.04 r16= 0.04; r17= 0.05; r18= 0.06; r19= 0.08; r110= 0.31 αterrestrial= 1 −Sum(r1i, i = 1, 10)

10

c Emittance of the surface at 350 C. Need to use radiation tables to generate the intervals. Using 10 points from Table 3.6.1b. The appropriate wavelenghts are: 3.02, 3.93, 4.65, 5.38, 6.15, 7.08, 8.23, 9.87, 12.6, 20.1. Reading reflectance values from Figure 4.8.3

r21= 0.59; r22= 0.72; r23= 0.77; r24= 0.84; r25= 0.87

r26= 0.90; r27= 0.91; r28= 0.93; r29= 0.94; r210= 0.94 Estimated r2[9] and r2[10]

 = 1 − Sum(r2i, i = 1, 10) 10

Solution

αextraterrestrial= 0.923 αterrestrial= 0.926  = 0.159

98

p04 04

Equations

4.4 Calculation of emittance for a real surface at two different temperatures.

$UnitSystem SI Mass Deg kPa K kJ Data

Ta = ConvertTemp(C, K, 175) Tb= ConvertTemp(C, K, 30)

a At 175 C

Use the 10 equal increments( from either Table 3.6.1b or as calculated here), and from the reflectance curve read the reflectances at the characteristic (midpoint) wavelengths. Note that it was necessary to limit λ between 0 and 100.

duplicate i = 1, 10 fa,i= i − 0.5

10

fa,i= BlBodyF ract (λa,i· Ta) end

Reading reflectances from curve C of Figure 4.8.2 are:

r1₁= .32; r1₂= .63; r1₃= .66; r1₄= .81; r1₅= .84; r1₆= .87; r1₇= .88 r18= .90; r19= .92; r110= .88

175= 1 − Sum(r1i, i = 1, 10) 10

b At 30 C

Repeat with new temperature.

duplicate i = 1, 10 fb,i= i − 0.5

10

fb,i= BlBodyF ract (λb,i· Tb) end

The corresponding reflectances from curve C of Figure 4.8.2 are:

r21= .69; r22= .82; r23= .87; r24= .88; r25= .90; r26= .91; r27= .92 r2₈= .91; r2₉= .88; r2₁₀= .87 estimated

30= 1 − Sum(r2i, i = 1, 10) 10

Note that it would have been useful to read the reflectances from the curve in small increments and place the values in a lookup table. Software exists to automatically digitize curves.

Solution

₁₇₅= 0.229 ₃₀= 0.135 T_a = 448.1[K] T_b= 303.2[K]

Arrays

Row f_a,i λ_a,i r1_i f_b,i λ_b,i r2_i [-] [micrometer] [-] [-] [micrometer] [-]

1 0.05 4.21 0.32 0.05 6.22 0.69

2 0.15 5.46 0.63 0.15 8.07 0.82

3 0.25 6.47 0.66 0.25 9.56 0.87

4 0.35 7.47 0.81 0.35 11.04 0.88

5 0.45 8.56 0.84 0.45 12.65 0.9

6 0.55 9.84 0.87 0.55 14.54 0.91

7 0.65 11.45 0.88 0.65 16.93 0.92

8 0.75 13.72 0.9 0.75 20.28 0.91

9 0.85 17.51 0.92 0.85 25.89 0.88

10 0.95 27.80 0.88 0.95 41.10 0.87

100

P04 05

Equations

4.5 Calculation of the α/ ratio for a surface in space.

$UnitSystem SI Mass Deg kPa K kJ Data

_λ,1= 0.20; _λ,2= 0.75; _λ,3= 0.90 (1)

λ1= 0.6; λ2= 2.6; λ3= 100 (2)

αλ,1= 1 − λ,1 (3)

αλ,2= 1 − λ,2 (4)

αλ,3= 1 − λ,3 (5)

T = 350 [K] (6)

From Table 1.3.1 for solar radiation the fraction in each band is:

fsolar,1= 0.351 (7)

fsolar,2= 0.968 + 0.2 · (0.981 − 0.968) (8)

fsolar,3= 1 (9)

∆f_solar,1= 0.351 (10)

∆f_solar,2= fsolar,2− fsolar,1 (11)

∆f_solar,3= f_solar,3− fsolar,2 (12)

From Eqn 4.6.4

α = αλ,1· ∆f_solar,1+ αλ,2· ∆f_solar,2+ αλ,3· ∆f_solar,3 (13) The fraction of the infrared radiation from 0 to λ is found from Table 3.6.1a or from the EES SETP function.

fIR,1= BlBodyF ract (λ1· T ) (14)

fIR,2= BlBodyF ract (λ2· T ) (15) f_IR,3= BlBodyF ract (1000 [micrometer] · T ) Note that we extended the data beyond the 100 micrometer limit.(16)

∆f_IR,1= f_IR,1 (17)

∆f_IR,2= f_IR,2− f_IR,1 (18)

∆f_IR,3= fIR,3− fIR,2 Note that almost all of the energy is in band 3. (19)

From Eqn 4.6.1

 = λ,1· ∆f_IR,1+ λ,2· ∆f_IR,2+ λ,3· ∆f_IR,3 (20)

Ratio

Ratio = α/ (21)

Solution

α = 0.4386 αλ,1= 0.8 αλ,2= 0.25 αλ,3= 0.1

∆fIR,1= 9.091 × 10⁻²⁶[-] ∆fIR,2= 0.0001006[-] ∆fIR,3= 0.9996[-] ∆fsolar,1= 0.351[-]

∆fsolar,2= 0.6196[-] ∆fsolar,3= 0.0294[-]  = 0.8997 λ,1= 0.2

λ,2= 0.75 λ,3= 0.9 fIR,1= 9.091 × 10⁻²⁶ fIR,2= 0.0001006 fIR,3= 0.9997 fsolar,1= 0.351[-] fsolar,2= 0.9706[-] fsolar,3= 1[-]

λ1= 0.6[micrometer] λ2= 2.6[micrometer] λ3= 100[micrometer] Ratio = 0.4875 T = 350[K]

2 102

P04 06

Equations

4.6 Calculation of solar absorptance as a function of angle of incidence.

function alpha/alpha_n(θ) (1)

Equation 4.11.1

alpha/alpha_n= 1 − 1.5879 × 10⁻³· θ + 2.731 × 10⁻⁴· θ²− 2.3026 × 10⁻⁵· θ³+ 9.0244 × 10⁻⁷· θ⁴

− 1.8000 × 10⁻⁸· θ⁵+ 1.7734 × 10⁻¹⁰· θ⁶− 6.9937 × 10⁻¹³· θ⁷ (2)

end (3)

Use 10 equally spaced intervals from Table 2.6.1. The appropriate wavelenghts (i.e., midpoints) are:

0.426, 0.508, 0.581, 0.653, 0.732, 0.822, 0.929, 1.080, 1.300, 1.9745. Reading reflectance values at these wavelengths for curve C of Figure 4.8.2:

ρ₁= 0.04; ρ₂= 0.08; ρ₃= 0.10; ρ₄= 0.07; ρ₅= 0.04 (4)

ρ₆= 0.10; ρ₇= 0.14; ρ₈= 0.22; ρ₉= 0.28; ρ₁₀= 0.19 (5)

From Eqn 4.6.4

αn= 1 − Sum(ρi, i = 1, 10)

10 (6)

Use Equation 4.11.1 to find α at angles of 30, 45 and 60 degrees.

α₃₀= α_n· alpha/alphan(30) (7)

α₄₅= α_n· alpha/alpha_n(45) (8)

α₆₀= α_n· alpha/alpha_n(60) (9)

Solution

Variables in Main program

α30= 0.86[-] α45= 0.85[-] α60= 0.81[-] αn = 0.874 ρ[1] = 0.04 ρ[2] = 0.08 ρ[3] = 0.1 ρ[4] = 0.07 ρ[5] = 0.04 ρ[6] = 0.1 ρ[7] = 0.14 ρ[8] = 0.22 ρ[9] = 0.28 ρ[10] = 0.19

P04 07a

Equations

4.7 Determine collector solar and IR radiation properties. This problem is best done in 4 parts.

function P lanck (λ, T ) (1)

C₁= C1#; C₂:= C2# (2)

P lanck₌ C₁

λ⁵· exp _λ·T^C2
− 1
(3)

end P lanck (4)

Subprogramemittance (T : ) (5)

T otal = sigma# · T⁴ (6)

 =

R100 [micrometer]

0.01 [micrometer]P lanck (λ, T ) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ

T otal (7)

104

end emittance (8)

a What is the absorptance for solar radiation assuming the sun is a blackbody at 5777?

T_sun= 5777 [K] (9)

T otal = sigma# · T_sun⁴ (10)

wavelength range 0.01 to 100 E0.01to100=

Z 100 0.01

P lanck (λ, Tsun) · Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = λ) dλ (11)

αsolar,0.01to100= 1 − E0.01to100/total (12)

wavelength range 0.041 to 25 E0.041to25=

Z 25 0.041

P lanck (lambda1, Tsun)·Interpolate(‘Curve C Fig 4 8 2’ , ‘lambda’ , ‘rho’ , ‘lambda’ = lambda1) dLambda1(13)

αsolar,0.041to25= 1 − E0.041to25/total (14)

The energy beyond 25 micrometers is very small. The energy below 0.41 micrometers is about 14% of the total. It isfortuitous that the two integrals give the same value to 3 significant figures.

Solution

18 1.454 0.2819

Plot 1

P04 07b

Equations

4.7 Determine collector solar and IR radiation properties. This problem is best done in 4 parts.

function P lanck (λ, T ) (1)

C1= C1#; C2:= C2# (2)

P lanck₌ C1

λ⁵· exp _λ·T^C2
− 1
(3)

end P lanck (4)

Subprogramemittance (T : ) (5)

T otal = sigma# · T⁴ (6)

b. Repeat problem a but assume the solar radiation is distributed as in Table 2.6.1.

We will use 10 increments. The variable limit are the bounds of the bands from Table 2.6.1.

limit1= .28; limit2= .47; limit3= .545; limit4= .617; limit5= .691; limit6= .777 (9)

Part a used a black body sun to estimate the solar absorptance as 0.887 whereas using the solar distribution of Table 2.6.1 the solar absorptance is estimated to be 0.876.

Part of this difference can be attributed to the different methods applied for integration.

Solution

Variables in Main program α_solar,b= 0.876

108

P04 07c

Equations

4.7 Determine collector solar and IR radiation properties. This problem is best done in 4 parts.

function P lanck (λ, T ) (1)

Subprogramemittance (T : ) (5)

T otal = sigma# · T⁴ (6)

c What is the emittance for long-wave radiation at 325 K?

TIR= 325 [K] (9)

T otalIR= sigma# · T_IR⁴ (10)

T $ = ‘Curve C Fig 4 8 2’ (11)

wavelength range 0.01 to 100 IR0.01to100=

Z 100 0.01

P lanck (lambda2, TIR) · Interpolate(T $, ‘lambda’ , ‘rho’ , ‘lambda’ = lambda2) dLambda2(12)

IR,0.01to100 = 1 − IR_0.01to100/total_IR (13)

wavelength range 0.041 to 25 IR0.041to25=

Z 25 0.041

P lanck (lambda3, TIR)·Interpolate(T $, ‘lambda’ , ‘rho’ , ‘lambda’ = lambda3) dLambda3(14)

IR,0.041to25 = 1 − IR_0.041to25/total_IR (15)

There is a very large difference between the two IR emittance values (0.376 vs 0.262) since about 14% of the energy is beyond 25 micrometers.

Solution

Variables in Main program

IR,0.01to100 = 0.3755 IR,0.041to25 = 0.2615 IR0.01to100= 395W/m²

IR0.041to25= 467.1W/m² Lambda2 = 100[micrometer] Lambda3 = 25[micrometer] T $ = ‘Curve C Fig 4 8 2’ T otalIR= 632.5W/m² TIR= 325[K]

110

P04 07d

Equations

4.7 Determine collector solar and IR radiation properties. This problem is best done in 4 parts.

function P lanck (λ, T ) (1)

C1= C1#; C2:= C2# (2)

P lanck₌ C1

λ⁵· exp _λ·T^C2
− 1
(3)

end P lanck (4)

Subprogramemittance (T : ) (5)

T otal = sigma# · T⁴ (6)

d Plot the emittance versus surface temperature for a range of temperatures from 300 K to 800 K.

Since EES has a limit on the number of integration variables that can be in the main program, an integral function is used.

duplicate k = 1, 11 (9)

TK= 250 + k · 50 (10)

call Emittance (TK : k) (11)

end (12)

A second order curve fit does not fo a very good job. A third order does much better

epsilon[i]=-0.51218632 + 0.00637126087*T[i] - 0.0000106300682*T[i]²+ 5.39253381E-09*T[i]³

Solution

6 0.6689 550 7 0.6463 600 8 0.6205 650 9 0.5932 700 10 0.5655 750 11 0.5383 800

epsilon vs T

112

P05 01

Equations

Calculate the reflectance of one glass surface.

function r (n, θ1) (1)

P05 02

P05 03

Equations

Transmittance of two covers at two angles.

function T ransIncAng (N cov, IncAng, KL, Ref rInd) (1)

Ref rAng = arcsin sin (IncAng) Ref rInd sin (Ref rAng + IncAng)

2

(6)

r_para= tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)

2

K = 20[1/m] L = 0.002[mm] N = 2 Rindex = 1.526 τa= 0.7814 τb= 0.7446 θa = 0[deg] θb= 50[deg]

116

P05 04

Equations

τ -α product for two covers at two angles.

We will use SETP functions TransIncAng and Absn.

Knowns

N = 2; θ_a= 25; θ_b= 60; KL = 0.037; α_n= 0.96; _p= 0.96; index = 1.526 (1) From footnote 2 of Section 5.5

ρd= 0.22 (2)

Part a

τ_a= T ransIncAng (N, θ_a, KL, index) (3)

α_a= α_n· Abs/Abs_n(θ_a) (4)

T auAlphaP roduct_a= τa· αa

(1 − (1 − α_a) · ρ_d) Eqn 5.5.1 (5)

Part b

τb= T ransIncAng (N, θb, KL, index) (6)

α_b= α_n· Abs/Absn(θ_b) (7)

T auAlphaP roduct_b= τ_b· α_b

(1 − (1 − αb) · ρd) Eqn 5.5.1 (8)

Solution

αa= 0.9475 αb= 0.8922 αn = 0.96 p= 0.96

index = 1.526 KL = 0.037 N = 2 ρd= 0.22

T auAlphaP roduct_a= 0.750 T auAlphaP roduct_b= 0.634 τa = 0.7828 τb = 0.6936

θa = 25[deg] θb= 60[deg]

P05 05

Equations

Estimate transmittance for diffuse sky and ground reflected radiation

We will use the SETP functions IncAngEffGrRef, IncAngEffDiff and TransIncAng Knowns

N = 1; KL = 0.037; index = 1.526; slope_a= 45; slope_b= 90 (1)

Part a

angle_g,a= IncAngEf f GrRef (Slope_a) (2)

angle_d,a= IncAngEf f Dif f (Slope_a) (3)

τg,a= T ransIncAng N, angle_g,a, KL, index

(4) τd,a= T ransIncAng N, angle_d,a, KL, index

(5)

Part a

angle_g,b= IncAngEf f GrRef (Slope_b) (6)

angle_d,b= IncAngEf f Dif f (Slope_b) (7)

τg,b= T ransIncAng N, angle_g,b, KL, index

(8) τd,b= T ransIncAng N, angle_d,b, KL, index

(9)

Solution

angle_d,a= 56.49[deg] angle_d,b= 59.33[deg] angle_g,a= 69.45[deg] angle_g,b= 59.81[deg] index = 1.526

KL = 0.037 N = 1 slope_a = 45 slope_b= 90 τd,a= 0.83

τd,b= 0.81 τg,a= 0.70 τg,b= 0.81

118

P05 06

Equations

function T ransIncAng (N cov, IncAng, KL, Ref rInd) (1)

Ref rAng = arcsin sin (IncAng) Ref rInd sin (Ref rAng + IncAng)

2

(6)

rpara= tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)

²

P05 07

Equations

Calculate transmittance and compare exact and approximate methods.

function T ransIncAng (N cov, IncAng, KL, Ref rInd) (1)

Ref rAng = arcsin sin (IncAng) Ref rInd sin (Ref rAng + IncAng)

2

(6)

r_para= tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)

²

Part a - single surface

Part b using the approximate method

τ1= T ransIncAng (N1, θ, K · L, index) (20)

Part c using the approximate method

τ₂= T ransIncAng (N₂, θ, K · L, index) (21)

Part d - calculate single cover transmittance by the exact method τ_Abs= exp

The two methods of calculation differ by 0.001 out of 0.808 or a little more than 0.1%

Solution

Variables in Main program

index = 1.526 K = 25 L = 0.0025 N1= 1 N2= 2 r = 0.074

r_{P ar} = 0.0003372 r_{P erp}= 0.1472 τ₁= 0.809 τ_1Exact= 0.808 τ₂= 0.686 τ_Abs= 0.9286 τP ar= 0.928 τP erp= 0.6883 θ = 55[deg] θ2= 32.47[deg]

P05 08

Equations

Estimate absorbed radiation for an inclined collector.

function T ransIncAng (N cov, IncAng, KL, Ref rInd) (1)

Ref rAng = arcsin sin (IncAng) Ref rInd sin (Ref rAng + IncAng)

²

(6)

rpara= tan (Ref rAng − IncAng) tan (Ref rAng + IncAng)

2

For the beam radiataion

τb= T ransIncAng (N, θb, KL, index) (17)

From Figure 5.4.1 at 45 degrees or Eqn 5.4.2

θd = 57 (18)

τd= T ransIncAng (N, θd, KL, index) (19)

From Figure 5.4.1 at 45 degrees or Eqn 5.4.1

θg = 69 (20)

P05 09

Equations

Estimate absorbed radiation for a collector.

We will use EES SETP functions TransIncAng, Abs_n, and I_{DIF F} . Knowns

Lat = 48; β = 63; θb= 36.2; n = 1; Index = 1.526; KL = 0.013; I = 1.03 MJ/m²

(1)

I_o= 1.58 MJ/m² ; Zenith = 71.5; ρ_g= 0.4; α_n= 0.955 (2)

a For beam radiation

τb= T ransIncAng (N, θb, KL, index) (3)

αb= αn· Abs/Absn(θb) (4)

Need diffuse reflectance of the cover for use in Equation 5.5.1. The equivalent diffuse angle from Figure 5.4.1 is 59 degrees.

Note that this is NOT the same angle as required for the diffuse radiation in part b.

τ_d59= T ransIncAng (N, 59, KL, index) (5)

T auAlpha_d59= τd59

T ransIncAng (n, 59, KL, 1) (6)

5.3.4

ρd59= T auAlphad59− τd59 (7)

5.3.6

T auAlpha_b= τ_b· α_b 1 − (1 − αb) · ρd59

(8)

Using Equation 5.5.2, TauAlphab=1.01*τb*αb=0.853 rather than 0.845

124

b For the diffuse radiation

The equivalent angle for diffuse sky radiation on a collector sloped at an angle of 63 degrees is 57 from Fig 5.4.1.

θd = 57 (9)

τd= T ransIncAng (n, θd, KL, index) (10)

α_d= α_n· Abs/Abs_n(θ_d) (11)

T auAlpha_d= τ_d· αd

1 − (1 − α_d) · ρ_d59 (12)

Using Equation 5.5.2, TauAlphad=1.01*τd*αd=0.772 rather than 0.766 c For the ground reflected radiation

The equivalent angle for ground reflected radiation on a collector sloped at an angle of 63 degrees is 64 from Fig 5.4.1.

θg = 64 (13)

d The absorbed solar radiation

kT = I/Io (17)

I_d= I · I_{DIF F /I}(k_T) ; I_b = I − I_d (18)

Rb= Cos(θb)

Cos(Zenith) (19)

Using Equation 5.9.1, the diffuse sky model:

Sb= T auAlpha_b· Ib· Rb (20)

Using Equation 5.9.2, the HDKR sky model, the results are S=1.99 rather than 1.75. This difference results from the choice of sky models and more than offsets ifferences resulting from choosing Equation 5.5.2 to evaluate TauAlpha.

A_i= I_b/I_o (24)

αb= 0.937 αd= 0.9009 αg= 0.8642 αn= 0.955

Ai= 0.4369 β = 63[deg] f = 0.8186 I = 1.03MJ/m²

Index = 1.526 Ib= 0.6903MJ/m²

Id= 0.3397MJ/m²

Io= 1.58MJ/m²

KL = 0.013 kT = 0.652 Lat = 48[deg] n = 1

ρd59= 0.021 ρg= 0.4 Rb= 2.543 S = 1.75MJ/m²

Sh = 1.99MJ/m²

Sb= 1.48MJ/m²

Sbh= 1.80MJ/m²

Sd= 0.1891MJ/m² Sdh= 0.12MJ/m²

Sg= 0.07734MJ/m²

Sgh= 0.08MJ/m²

T auAlpha_b= 0.843 T auAlpha_d= 0.766 T auAlpha_d59= 0.857 T auAlpha_g= 0.6876 τb= 0.899

τ_d= 0.848 τ_d59= 0.836 τ_g= 0.7939 θ_b= 36.2[deg]

θ_d = 57[deg] θ_g= 64[deg] Zenith = 71.5[deg]

126

P05 10

Equations

Determine a months absorbed radiation.

We will use EES SETP functions TransIncAng, Abs/Absnand HDIF F BAR/HBAR.

Knowns

Lat = 43; slope = 90; n = 1; , Index = 1.526; KL = 0.013; ρg= 0.6; αn= 0.93 (1) (From Appendix G)

H = 6.44;¯ K¯t= 0.47 (2)

First need to find average for the three components.

For beam radiation:

θb = 40 (3)

τb= T ransIncAng (n, θb, KL, Index) (4)

αb= αn· Abs/Absn(θb) (5)

Need ρ_dof the cover for use in Equation 5.5.1. The equivalent diffuse angle from Figure 5.4.1 is 59 degrees -&

this is generally NOT the same angle as required for the diffuse sky radiation.

τ_d59= T ransIncAng (n, 59, KL, Index) (6)

T auAlpha_d59= τd59

T ransIncAng (n, 59, KL, 1) Eqn 5.3.4 (7)

ρd59= T auAlphad59− τd59 Eqn 5.3.6 (8)

T auAlpha_b= τb· αb

1 − (1 − α_b) · ρ_d59 (9)

For the diffuse radiation:

θd = 59 (10)

The equivalent angle for diffuse sky radiation on a vertical collector is 59 from Fig 5.4.1 -the same angle as used in some calculations in part a.

τd= τd59 (11)

αd= αn· Abs/Absn(θd) (12) T auAlpha_d= τ_d· α_d

1 − (1 − αd) · ρd59

(13)

For the ground reflected radiation:

From Fig 5.4.1 the equivalent angle for ground reflected radiation on a vertical collector is exactly the same as for diffuse sky radiation.

T auAlpha_g= T auAlpha_d (14)

Need to break total into beam and diffuse - use Erbs’ correlation for winter - i.e. ω_sis less than 81.4 - use 50 H¯d= ¯H · Hdif f Bar/HBar Lat, n, ¯KT

(15)

H¯b= ¯H − ¯Hd (16)

R¯b= 2.62 From Figure 2.19.1d (17)

a) The absorbed solar radiation components using Equation 5.10.2, the diffuse sky model.

a) The absorbed solar radiation components using Equation 5.10.2, the diffuse sky model.

In document DE EMPRESAS (página 33-45)